Novice Astronomy / Physics Question, wrt spectroscopes.

I found myself perplexed recently by spectroscopes, and would
appreciate it if someone wiser than I out there would clear up my
confusion. Probably this stuff is well understand and not deserving
of being mentioned here; if you could then just provide me with a few
good words for search terms to find relevant data it'd be great.

As I understand it, light is emitted by things as electrons in excited
energy levels fall down into lower, standard energy levels. This
nicely explains why looking at many things through a spectroscope
reveals a few bright wavelengths of light in a sea of darkness.

It makes it non-obvious, however, why the sun produces a large, mostly
continuous band of colors. One site suggested that this occurs in
heated gases of high pressure, while the former phenomenon occurs in
heated gases of low pressure, but supplied no reasoning, and this
behavior is not self-evident to me.

My last point of confusion (to be mentioned here!) surrounds the gaps
in the "mostly continuous band of colors" mentioned above. As I
understand it, these gaps represent absorbed / scattered wavelengths
of light that correspond to the wavelengths of light emitted by heated
gases as discussed two paragraphs above. This makes sense; I just
don't understand how a _single_ wavelength of light's being missing
could make an overall difference in the spectrograph's shown spectrum.
It would seem to me that the infinitely(?) large number of
wavelengths produced would make distinguishing a single one
impossible, much as seeing a single missing, infinitely small point on
a line would not be possible. Can an electron of an element's atom be
excited by a whole, albeit very small, range of wavelengths / energies
of light, or just a single one that composes an infinitely small
portion of the overall spectrum?

In other words, if the dark line represents only one wavelength, how
is this gap appreciable in a spectrum consisting of an infinite(?)
number of wavelengths?

Thanks for your help. Again, if these questions are standard fare,
just throw some keywords at me and I'll try searching the internet
again...

E Schlafly

In article , (E Schlafly) wrote:

snip lots of good questions

You're actually asking questions which would be covered in a fairly broad
university course in astronomical spectroscopy. I don't think all of your
questions can be answered in one go, because there's a lot of physics to
learn first.

One thing to remember is that although atoms tend to emit or absorb at
particular wavelengths, in a hot source such as the sun, the atoms are
moving around quickly which will broaden the emission or absorption lines as
seen on the earth. Also, you're seeing several transitions from several
species of atoms and molecules, as well as ionized atoms and molecules,
which will tend to give you a broad continuum. In addition, you also have
simple thermal emission (i.e., blackbody radiation).

In other words, if the dark line represents only one wavelength, how
is this gap appreciable in a spectrum consisting of an infinite(?)
number of wavelengths?

I'm not quite sure what you're asking. Generally, when doing spectroscopy,
you look at one particular part of the spectrum, not the entire
electromagnetic spectrum. Now, given that, if you're looking for a spectral
line of a particular atom, ion or molecule, you look at a region of the
spectrum that this species will absorb or emit light. If it's there and the
abundance is high enough, you'll see the line or band against the general
continuum. Of course, there may be other lines from other atoms or molecules
present, and this is why astronomical spectroscopy can be very hard. You
also need to use a high enough resolution to distinguish the line from the
continuum. At low resolution, a sharp line will virtually blend into the
continuum, whereas at high resolution, you can actually distinguish the
line. It all depends on what you're trying to detect. For instance, atomic
lines are generally quite sharp, so you need high resolution to see them.
Molecular bands, however, are generally quite broad, so you can see them at
lower resolution. Then again, to actually see what's going on in a molecular
band, you need high resolution to distinguish between the various bands
(e.g., a molecule will have both vibrational and rotational modes).

Thanks for your help. Again, if these questions are standard fare,
just throw some keywords at me and I'll try searching the internet
again...

As well as looking up spectroscopy and astronomy, you could add
"pressure broadening", "solar spectrum", "continuum", "molecular", "atomic".
In fact, there are a whole host of things you could add (not all at the same
time of course!) which I can't think of straight off the top of my head.

In article , E Schlafly wrote:
[...]
In other words, if the dark line represents only one wavelength, how
is this gap appreciable in a spectrum consisting of an infinite(?)
number of wavelengths?

I'll supplement Tom Kerr's comments. This is a good question,
the answer being that spectral lines do have finite width
(range of wavelengths) which depends on the conditions they
formed under -- like the gas temperature (doppler shift from gas motion),
gas pressure (interactions with other atoms shift the energy levels),
and the amount of time the atoms spend in their less- and
more-excited states (quantum uncertainty in energy!).
A lot of spectroscopy seems to go into study of detailed line profiles.

If a line really did have zero width, you're right, it would
obscure an undetectably small amount of continuum light.

Stuart Levy

E Schlafly wrote:
I found myself perplexed recently by spectroscopes, and would
appreciate it if someone wiser than I out there would clear up my
confusion. Probably this stuff is well understand and not deserving
of being mentioned here; if you could then just provide me with a few
good words for search terms to find relevant data it'd be great.


[rest deleted for brevity]

Your rambling style makes it difficult to try to get to whatever question(s) you
might have been asking, so this response may not provide the answer. The
following comes from a college textbook I have used in teaching classes in the
past and addresses the continuum produced in the Sun's spectrum:

"The mechanism that causes continuous emission is different for different
spectral types (of stars - my clarification). In the Sun and other stars of
similar similar spectral types, the continuous emission is caused by a strange
type of ion: the negative hydrogen ion. An ordinary ion has one or more
electrons missing, and so has a positive charge and could be called a positive
ion. A negative ion, on the other hand, has one or more extra electrons. Only
a tiny fraction of the hydrogen atoms are in such a state, with one extra
electron, but the Sun is so overwhelmingly (90 percent) hydrogen that enough
negative hydrogen atoms are present to be important.

"When a neutral hydrogen atom takes up an extra electron, the second electron
may temporarily become part of the atom, residing in an energy level. But the
energy difference between the extra electron's energy and its energy as part of
the atom are not limited to discrete values, because the electron could have had
any amount of energy before it joined the atom. Alternatively, the extra
electron stars free of the atom, is affected by the atom, but winds up still
free of the atom. In this case, too, the amount of energy involved is not
limited to discrete values. Thus a continuous spectrum is formed. The case is
thus different from the one in which an electron in an atom jumps between two
energy levels, because the electron has a fixed amount of energy in each level,
and a fixed number subtracted from another fixed number gives a fixed number.
Subtracting one fixed number from another cannot lead to a continuous range of
values." ("Astronomy: From the Earth to the Universe", 4th edition, by Jay M.
Pasachoff. p. 344).

He goes on with examples and further comments for another page, so if you are
interested, you might see if you can find the mentioned text or its later
editions at a local college bookstore. But, once the continuum is in place, the
"normal" gases can then react to the available radiation, absorbing that
significant for the electron energy states, removing that radiation, then
reradiating it in all possible directions, effectively removing enough of it in
the direction of our detectors that we see dark lines from those various gases.

Hope that gets to what you were asking.

Thank you for all of your helpful responses (and sorry for rambling).
I am now temporarily less confused.

E Schlafly

In article , (E Schlafly) wrote:
Thank you for all of your helpful responses (and sorry for rambling).
I am now temporarily less confused.

Thanks for saying "thanks"! It's pretty common on usenet for someone to ask
questions and then not acknowledge the responses (I'm guilty of it myself).

You will probably get more useful help in the future if you can ask more
specific questions. I think the answers you got showed that you will get
good help, but that your original questions were too vague to answer in a
concise way.

In article ,
(E Schlafly) writes:
As I understand it, light is emitted by things as electrons in excited
energy levels fall down into lower, standard energy levels. This
nicely explains why looking at many things through a spectroscope
reveals a few bright wavelengths of light in a sea of darkness.

These are really good questions. There have already been some good
answers, and I'll just add a couple of thoughts.

It makes it non-obvious, however, why the sun produces a large, mostly
continuous band of colors. One site suggested that this occurs in
heated gases of high pressure, while the former phenomenon occurs in
heated gases of low pressure, but supplied no reasoning, and this
behavior is not self-evident to me.

One good way to look at things is to consider "brightness
temperature." I'm not going to define it, but higher temperature
means the light is brighter. When an object is in "local
thermodynamic equilibrium," which I'm also not going to define, a
good approximation to the object's brightness at any wavelength is to
imagine shining a photon of that wavelength towards the object.
Figure out at what layer in the object the photon would be absorbed,
then find out the temperature of that layer. That will be the
brightness temperature of the _emerging_ radiation from the object.

For stars, all incoming light is absorbed near the surface, because
the gas density increases very rapidly below the surface. Thus
brightness temperature is about the same as the surface temperature.
However, in the centers of the atomic lines, the light is absorbed
more strongly and hence an incoming "line photon" would be absorbed
at a height _above_ where an incoming "continuum photon" would be
absorbed. Since the temperature of stars generally increases with
depth, the continuum will have a higher brightness temperature than
the lines. Thus the lines will be "in absorption."

There is an exception for the very strongest lines. In the Sun's
atmosphere, there is a temperature minimum, and temperature increases
above that height as well as below it. At the wavelength of a very
strong line, an incoming photon would be absorbed above the
temperature minimum, and the line is seen "in emission." In the Sun,
emission lines are common in the UV but rare in visible light. (I
don't think there are any, although some of the lines have little
emission peaks in their centers.)

All this is a very simplified explanation, but I think it may give
some feeling for what is going on. Unfortunately, I don't know a
good source for an explanation of radiative transfer at a
non-technical level.

I just don't understand how a _single_ wavelength of light's being
missing could make an overall difference in the spectrograph's
shown spectrum.

This is good thinking. As others have said, all lines have a finite
width, which is determined by local temperature and pressure,
velocity gradients or turbulence, and also the intrinsic width of the
atomic line itself. Even for a completely isolated atom, completely
at rest, there is some finite range of wavelengths capable of
exciting a transition. (Of course it's a probability distribution,
not a sharp cuton/cutoff. That is, a photon at the exact center
wavelength has the highest probability of exciting a transition, but
other wavelengths have non-zero probability.)

Hope this helps. It's quite a pleasant change to see sensible
questions in this newsgroup.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)

ES
From: (E Schlafly)
ES
Subject: Novice Astronomy / Physics Question, wrt spectroscopes.
ES
Date: 27 Jul 2003 01:36:15 -0700
ES
Organization: http://groups.google.com/

You asked the same questions Kirchoff did in the mid 1800s! From
these questions he wroked up the three kinds of spectra: bright line,
dark line, and cintinuous. You explanation of the bright line
spectrunm is more or less correct. Start with that.
The continuous spectrum is a blackbody spectrum, first hinted at
by John Draper (homeboy from Greenwxih Village, Manhattan). The
thermal agitation of the gas atoms radiate at all wavelengths in a mix
given by Plank's rquation. This is the humpback curve or continuum you
see in spectrodensitometer tracings.
The Sun gives off, by its temperature, a blackbody continuum. That
radiation passes thru other gases above the Sun's surface. Certain of
the wavelengths in that blackbody radiation are the exact ones to
excite the atoms in this superincumbent gas. This gas accepts the
radiation at these peculiar wavelengths, And there by removes those
wavelengths from the continuum. Result is that above the Sun's outer
gases, the raidation consists of the continuum laced by dark lines at
the wavelengths knocked out of it.



ES I found myself perplexed recently by spectroscopes, and would
ES appreciate it if someone wiser than I out there would clear up my
ES confusion. Probably this stuff is well understand and not deserving
ES of being mentioned here; if you could then just provide me with a few
ES good words for search terms to find relevant data it'd be great.
ES
ES As I understand it, light is emitted by things as electrons in excited
ES energy levels fall down into lower, standard energy levels. This
ES nicely explains why looking at many things through a spectroscope
ES reveals a few bright wavelengths of light in a sea of darkness.
ES
ES It makes it non-obvious, however, why the sun produces a large, mostly
ES continuous band of colors. One site suggested that this occurs in
ES heated gases of high pressure, while the former phenomenon occurs in
ES heated gases of low pressure, but supplied no reasoning, and this
ES behavior is not self-evident to me.
ES
ES My last point of confusion (to be mentioned here!) surrounds the gaps
ES in the "mostly continuous band of colors" mentioned above. As I
ES understand it, these gaps represent absorbed / scattered wavelengths
ES of light that correspond to the wavelengths of light emitted by heated
ES gases as discussed two paragraphs above. This makes sense; I just
ES don't understand how a _single_ wavelength of light's being missing
ES could make an overall difference in the spectrograph's shown spectrum.
ES It would seem to me that the infinitely(?) large number of
ES wavelengths produced would make distinguishing a single one
ES impossible, much as seeing a single missing, infinitely small point on
ES a line would not be possible. Can an electron of an element's atom be
ES excited by a whole, albeit very small, range of wavelengths / energies
ES of light, or just a single one that composes an infinitely small
ES portion of the overall spectrum?
ES
ES In other words, if the dark line represents only one wavelength, how
ES is this gap appreciable in a spectrum consisting of an infinite(?)
ES number of wavelengths?

---
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